Find all 2 by 2 complex matrix with the following condition












1















Find all $A in mathrm{Mat}_{2 times 2} ( mathbb{C})$ such that
$A^2 = -I$ and prove that there is no 2 by 2 real diagonal matrices
$A$ with $A^2 = -I$. Deduce that for every even $n$ there are
infinitely many $n times n$ real matrices with $A^2 = -I$.




Try:



We have



$$ A = begin{bmatrix} a & b \ c & d end{bmatrix} implies A^2 = begin{bmatrix}a^2+bc & b(a+d) \ c(a+d) & cb+d^2 end{bmatrix} $$



So $A^2 = - I$ iff $-1 = a^2+bc = cb + d^2$ and $0=b(a+d)=c(a+d)$



If $b=0$, then we have $a^2=-1 $ , $d^2=-1$.



if $a^2=-1$, then $a= pm i$ and similarly $b= pm i$. As long as $a+d neq 0$, we see that $c=0$ otherwise $c in mathbb{C}$ is arbitrary and so we have



$$ begin{bmatrix} i & 0 \ 0 & i end{bmatrix}, begin{bmatrix} -i & 0 \ 0 & -i end{bmatrix},begin{bmatrix} -i & 0 \ c & i end{bmatrix},begin{bmatrix} i & 0 \ c & -i end{bmatrix} $$



now if $b neq 0$, then $a+d=0$ and so $a=-d$ and thus $c = -(1+a^2)/b$



Therefore, we have



$$ begin{bmatrix} a & b \ -(1+a^2)/b & -a end{bmatrix} $$



are all posible matrices. Now, if the $a,b,c,d$ were real, we see that $b neq 0$ and so we get matrices like the one above only.



Now, is this correct so far? I'm stuck for the $ntimes n$ case. Any help?






linear-algebra







share|cite|improve this question





























    1















    Find all $A in mathrm{Mat}_{2 times 2} ( mathbb{C})$ such that
    $A^2 = -I$ and prove that there is no 2 by 2 real diagonal matrices
    $A$ with $A^2 = -I$. Deduce that for every even $n$ there are
    infinitely many $n times n$ real matrices with $A^2 = -I$.




    Try:



    We have



    $$ A = begin{bmatrix} a & b \ c & d end{bmatrix} implies A^2 = begin{bmatrix}a^2+bc & b(a+d) \ c(a+d) & cb+d^2 end{bmatrix} $$



    So $A^2 = - I$ iff $-1 = a^2+bc = cb + d^2$ and $0=b(a+d)=c(a+d)$



    If $b=0$, then we have $a^2=-1 $ , $d^2=-1$.



    if $a^2=-1$, then $a= pm i$ and similarly $b= pm i$. As long as $a+d neq 0$, we see that $c=0$ otherwise $c in mathbb{C}$ is arbitrary and so we have



    $$ begin{bmatrix} i & 0 \ 0 & i end{bmatrix}, begin{bmatrix} -i & 0 \ 0 & -i end{bmatrix},begin{bmatrix} -i & 0 \ c & i end{bmatrix},begin{bmatrix} i & 0 \ c & -i end{bmatrix} $$



    now if $b neq 0$, then $a+d=0$ and so $a=-d$ and thus $c = -(1+a^2)/b$



    Therefore, we have



    $$ begin{bmatrix} a & b \ -(1+a^2)/b & -a end{bmatrix} $$



    are all posible matrices. Now, if the $a,b,c,d$ were real, we see that $b neq 0$ and so we get matrices like the one above only.



    Now, is this correct so far? I'm stuck for the $ntimes n$ case. Any help?






    linear-algebra







    share|cite|improve this question



























      1












      1








      1








      Find all $A in mathrm{Mat}_{2 times 2} ( mathbb{C})$ such that
      $A^2 = -I$ and prove that there is no 2 by 2 real diagonal matrices
      $A$ with $A^2 = -I$. Deduce that for every even $n$ there are
      infinitely many $n times n$ real matrices with $A^2 = -I$.




      Try:



      We have



      $$ A = begin{bmatrix} a & b \ c & d end{bmatrix} implies A^2 = begin{bmatrix}a^2+bc & b(a+d) \ c(a+d) & cb+d^2 end{bmatrix} $$



      So $A^2 = - I$ iff $-1 = a^2+bc = cb + d^2$ and $0=b(a+d)=c(a+d)$



      If $b=0$, then we have $a^2=-1 $ , $d^2=-1$.



      if $a^2=-1$, then $a= pm i$ and similarly $b= pm i$. As long as $a+d neq 0$, we see that $c=0$ otherwise $c in mathbb{C}$ is arbitrary and so we have



      $$ begin{bmatrix} i & 0 \ 0 & i end{bmatrix}, begin{bmatrix} -i & 0 \ 0 & -i end{bmatrix},begin{bmatrix} -i & 0 \ c & i end{bmatrix},begin{bmatrix} i & 0 \ c & -i end{bmatrix} $$



      now if $b neq 0$, then $a+d=0$ and so $a=-d$ and thus $c = -(1+a^2)/b$



      Therefore, we have



      $$ begin{bmatrix} a & b \ -(1+a^2)/b & -a end{bmatrix} $$



      are all posible matrices. Now, if the $a,b,c,d$ were real, we see that $b neq 0$ and so we get matrices like the one above only.



      Now, is this correct so far? I'm stuck for the $ntimes n$ case. Any help?






      linear-algebra







      share|cite|improve this question
















      Find all $A in mathrm{Mat}_{2 times 2} ( mathbb{C})$ such that
      $A^2 = -I$ and prove that there is no 2 by 2 real diagonal matrices
      $A$ with $A^2 = -I$. Deduce that for every even $n$ there are
      infinitely many $n times n$ real matrices with $A^2 = -I$.




      Try:



      We have



      $$ A = begin{bmatrix} a & b \ c & d end{bmatrix} implies A^2 = begin{bmatrix}a^2+bc & b(a+d) \ c(a+d) & cb+d^2 end{bmatrix} $$



      So $A^2 = - I$ iff $-1 = a^2+bc = cb + d^2$ and $0=b(a+d)=c(a+d)$



      If $b=0$, then we have $a^2=-1 $ , $d^2=-1$.



      if $a^2=-1$, then $a= pm i$ and similarly $b= pm i$. As long as $a+d neq 0$, we see that $c=0$ otherwise $c in mathbb{C}$ is arbitrary and so we have



      $$ begin{bmatrix} i & 0 \ 0 & i end{bmatrix}, begin{bmatrix} -i & 0 \ 0 & -i end{bmatrix},begin{bmatrix} -i & 0 \ c & i end{bmatrix},begin{bmatrix} i & 0 \ c & -i end{bmatrix} $$



      now if $b neq 0$, then $a+d=0$ and so $a=-d$ and thus $c = -(1+a^2)/b$



      Therefore, we have



      $$ begin{bmatrix} a & b \ -(1+a^2)/b & -a end{bmatrix} $$



      are all posible matrices. Now, if the $a,b,c,d$ were real, we see that $b neq 0$ and so we get matrices like the one above only.



      Now, is this correct so far? I'm stuck for the $ntimes n$ case. Any help?






      linear-algebra




      linear-algebra






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 4 at 6:01









      Lee

      14010




      14010










      asked Jan 4 at 5:52









      Jimmy SabaterJimmy Sabater

      1,980219




      1,980219






















          1 Answer
          1






          active

          oldest

          votes


















          2














          Looking very good so far. However, for the $a=pm i, d=-a$ case, you missed the possibility $c=0$ with $b$ being anything.



          For even $n$, take the real $2times2$ matrix you've found, make $frac n2$ copies of it all along the diagonal of an $ntimes n$ matrix (whether you make them all equal, or just make sure they are all of that form is up to you), and otherwise make all entries $0$. This easily gives you infinitely many different $ntimes n$ matrices.



          Bonus question: What happens for odd $n$ with real matrices?






          share|cite|improve this answer























            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061350%2ffind-all-2-by-2-complex-matrix-with-the-following-condition%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2














            Looking very good so far. However, for the $a=pm i, d=-a$ case, you missed the possibility $c=0$ with $b$ being anything.



            For even $n$, take the real $2times2$ matrix you've found, make $frac n2$ copies of it all along the diagonal of an $ntimes n$ matrix (whether you make them all equal, or just make sure they are all of that form is up to you), and otherwise make all entries $0$. This easily gives you infinitely many different $ntimes n$ matrices.



            Bonus question: What happens for odd $n$ with real matrices?






            share|cite|improve this answer




























              2














              Looking very good so far. However, for the $a=pm i, d=-a$ case, you missed the possibility $c=0$ with $b$ being anything.



              For even $n$, take the real $2times2$ matrix you've found, make $frac n2$ copies of it all along the diagonal of an $ntimes n$ matrix (whether you make them all equal, or just make sure they are all of that form is up to you), and otherwise make all entries $0$. This easily gives you infinitely many different $ntimes n$ matrices.



              Bonus question: What happens for odd $n$ with real matrices?






              share|cite|improve this answer


























                2












                2








                2






                Looking very good so far. However, for the $a=pm i, d=-a$ case, you missed the possibility $c=0$ with $b$ being anything.



                For even $n$, take the real $2times2$ matrix you've found, make $frac n2$ copies of it all along the diagonal of an $ntimes n$ matrix (whether you make them all equal, or just make sure they are all of that form is up to you), and otherwise make all entries $0$. This easily gives you infinitely many different $ntimes n$ matrices.



                Bonus question: What happens for odd $n$ with real matrices?






                share|cite|improve this answer














                Looking very good so far. However, for the $a=pm i, d=-a$ case, you missed the possibility $c=0$ with $b$ being anything.



                For even $n$, take the real $2times2$ matrix you've found, make $frac n2$ copies of it all along the diagonal of an $ntimes n$ matrix (whether you make them all equal, or just make sure they are all of that form is up to you), and otherwise make all entries $0$. This easily gives you infinitely many different $ntimes n$ matrices.



                Bonus question: What happens for odd $n$ with real matrices?







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 4 at 6:16

























                answered Jan 4 at 6:04









                ArthurArthur

                111k7105186




                111k7105186






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061350%2ffind-all-2-by-2-complex-matrix-with-the-following-condition%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    1300-talet

                    Image
                    Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
                    Read more

                    1300-talet

                    Image
                    Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
                    Read more

                    Display a custom attribute below product name in the front-end Magento 1.9.3.8

                    Image
                    0 I need to display a custom attribute bellow the product name and I don't know how can I call this attribute. I am editing the file /app/design/frontend/templatemela/MAG100219_1/template/catalog/product/view.phtml But how can I call this attribute in php? I tried calling a string echo with the name I put in the attribute, but it didn't work. Code is as below: <div class="page-title product-title"> <h1><?php echo $_helper->productAttribute($_product, $_product->getName(), 'name') ?></h1> </div> <div class="codigoinmetro"> <h2><?php echo $_helper->productAttribute($_product, $_product->getName(), 'registroinmetro') ?></h2> </div> php attrib...
                    Read more